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(C) Let $P$ be $(x_1, y_1)$.The equation of a circle of radius $r$ which touches the coordinate axes and lies in the first quadrant is $(x-r)^2 + (y-r

Vedclass · Accepted Answer

$(2r, 3r)$

Vedclass · Answer

(C) Let $P$ be $(x_1, y_1)$.The equation of a circle of radius $r$ which touches the coordinate axes and lies in the first quadrant is $(x-r)^2 + (y-r)^2 = r^2$,which simplifies to $x^2 + y^2 - 2xr - 2yr + r^2 = 0$.The polar of $P(x_1, y_1)$ with respect to the circle $x^2 + y^2 - 2xr - 2yr + r^2 = 0$ is given by $xx_1 + yy_1 - r(x+x_1) - r(y+y_1) + r^2 = 0$.Rearranging the terms,we get $(x_1-r)x + (y_1-r)y = r(x_1+y_1-r)$.Given that the polar is $x+2y=4r$,we compare the coefficients:$\frac{x_1-r}{1} = \frac{y_1-r}{2} = \frac{r(x_1+y_1-r)}{4r} = \frac{x_1+y_1-r}{4}$.From $\frac{x_1-r}{1} = \frac{y_1-r}{2}$,we get $2x_1 - 2r = y_1 - r$,so $y_1 = 2x_1 - r$.From $\frac{x_1-r}{1} = \frac{x_1+y_1-r}{4}$,we get $4x_1 - 4r = x_1 + y_1 - r$,so $3x_1 - 3r = y_1$.Equating the two expressions for $y_1$: $2x_1 - r = 3x_1 - 3r$,which gives $x_1 = 2r$.Substituting $x_1 = 2r$ into $y_1 = 2x_1 - r$,we get $y_1 = 2(2r) - r = 3r$.Thus,the point $P$ is $(2r, 3r)$.

If the polar of a point $P$ with respect to a circle of radius $r$ which touches the coordinate axes and lies in the first quadrant is $x+2y=4r$,then the point $P$ is

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